Functions

Definition

Let $F\subseteq A\times B$ so its a relation

Suppose F has the property that if $a\in A$ appears as a a first entry in an element of $F$ , then it is the only element of $F$ that it appears in

Then we call $F$ a function

If $(a,b)\in F$ , we write $f(a) = b$

In a function, second entries can be matched with more than one first entry

If, $F\subseteq A\times B$ is a function with $f(a)=b\ iff\ (a,b) \leftarrow F$ we often omit $F$ and only refer to $f$ and say:

$f:A\rightarrow B$
bdg-infoWe say $f$ maps $A$ into $B$

Consider the following example

A &= \Set{0,1,2,3,4,5,6,7,8,9}\\ B &= \Set{r,s,t,u,v,w,x,y,z}\\ F &= \Set{(0,r), (1,s), (2,s), (3,t), (4,u), (5,w), (6,x), (7,x), (8,x), (9,x) }

Then the rule for $f$ may be stated explicitly or inferred from $F$

bdg-warningNote $range(f) = \Set{r,s,t,u,v,w,x,y,z} \subset B$ is a proper subset of $B$

Let us write $f^{-1}(b)=$ pre-image of $b$ under $f$ $=\Set{a\in A\mid b = f(a)}$

For our example

$f^{-1}(z)=\emptyset$
$f^{-1}(t)=\Set{3},\ f^{-1}(x)=\Set{6,7,8,9}$

Composition of function

Let $f: A \rightarrow B$ & $g: B \rightarrow C$ be two function

We define the composition of $g$ and $f$ to be the function $g\circ f:A\rightarrow C$ defined by $g\circ f(a)=g(f(a))$

Composition of More Than 2 Function

We can define the composition of more than 2 functions if $f: A \rightarrow B$ , $g: B \rightarrow G$ , $h: G \rightarrow D$ , then

$h\circ g\circ f:A\rightarrow D$ and $h\circ g\circ f(a)=h(g(f(a)))$ etc

Inverse Functions

Let $f:A\rightarrow B$ be a 1-1 onto map

Then for $\forall b\in B, \exists\text{ exactly one }a\in A\ st\ f(a)=b$

We define the inverse function of $f$ , called $f^{-1}(b)=a$ iff $f(a)=b$ and we say $f$ is invertible

Identity Function

If we let $id_A:A\rightarrow A$ be the identity map on $A$ , $id_A(a)=a,\forall a\leftarrow A$ and $id_B:B\rightarrow B$ be the identity map on $B,\ id_B(b)=b$ , then for $f$ , 1-1, onto, $f^{-1}\circ f =id_A$ , $f\circ f^{-1}=id_B$

Note

If $f$ is not 1-1, then we can't define $f^{-1}$ , since if $f(a_1)=b$ , $f(a_2)=b$ , then $a_1\ne a_2$ bdg-dangernonsense because how do we define $f^{-1}(b)$ ? Is it $a_1$ or $a_2$ ?

If $f:A\rightarrow B$ is 1-1 but not onto, ie $f(a)\subset B$ , we can define $f^{-1}$ on $f(A) \subset B$ since $f:A\rightarrow f(A)$ is 1-1 and onto and since $f^(-1):f(A)\rightarrow A$ is a well defined function

Often we deal with $f: A \rightarrow A$ , then it is possible to compose two functions from $A$ to $A$ in different order, ie $f\circ g$ or $g\circ f$